Question: Simplify and expand the following expression: $ \dfrac{5}{3z + 15}- \dfrac{2}{z - 8}- \dfrac{2}{z^2 - 3z - 40} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{5}{3z + 15} = \dfrac{5}{3(z + 5)}$ We can factor the quadratic in the third term: $ \dfrac{2}{z^2 - 3z - 40} = \dfrac{2}{(z + 5)(z - 8)}$ Now we have: $ \dfrac{5}{3(z + 5)}- \dfrac{2}{z - 8}- \dfrac{2}{(z + 5)(z - 8)} $ The least common multiple of the denominators is: $ 3(z + 5)(z - 8)$ In order to get the first term over $3(z + 5)(z - 8)$ , multiply by $\dfrac{z - 8}{z - 8}$ $ \dfrac{5}{3(z + 5)} \times \dfrac{z - 8}{z - 8} = \dfrac{5(z - 8)}{3(z + 5)(z - 8)} $ In order to get the second term over $3(z + 5)(z - 8)$ , multiply by $\dfrac{3(z + 5)}{3(z + 5)}$ $ \dfrac{2}{z - 8} \times \dfrac{3(z + 5)}{3(z + 5)} = \dfrac{6(z + 5)}{3(z + 5)(z - 8)} $ In order to get the third term over $3(z + 5)(z - 8)$ , multiply by $\dfrac{3}{3}$ $ \dfrac{2}{(z + 5)(z - 8)} \times \dfrac{3}{3} = \dfrac{6}{3(z + 5)(z - 8)} $ Now we have: $ \dfrac{5(z - 8)}{3(z + 5)(z - 8)} - \dfrac{6(z + 5)}{3(z + 5)(z - 8)} - \dfrac{6}{3(z + 5)(z - 8)} $ $ = \dfrac{ 5(z - 8) - 6(z + 5) - 6} {3(z + 5)(z - 8)} $ Expand: $ = \dfrac{5z - 40 - 6z - 30 - 6}{3z^2 - 9z - 120} $ $ = \dfrac{-z - 76}{3z^2 - 9z - 120}$